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Chapter 6

Complex Analysis and Conformal Mapping

The term complex analysis refers to the calculus of complex-valued functions f(z)depending on a single complex variable z. To the novice, it may seem that this subjectshould merely be a simple reworking of standard real variable theory that you learnedin first year calculus. However, this nave first impression could not be further fromthe truth! Complex analysis is the culmination of a deep and far-ranging study of thefundamental notions of complex differentiation and integration, and has an elegance andbeauty not found in the real arena. For instance, complex functions are necessarily analytic,meaning that they can be represented as convergent power series and hence are infinitelydifferentiable. Thus, difficulties with degree of smoothness, strange discontinuities, deltafunctions, and other pathological properties of real functions never arise in the complexrealm.

The driving force behind many of the applications of complex analysis is the remark-able connection between harmonic functions of two variables a.k.a. solutions of theplanar Laplace equation and complex functions. To wit, the real and imaginary partsof any complex analytic function are automatically harmonic. In this manner, complexfunctions provide a rich lode of additional solutions to the two-dimensional Laplace equa-tion, which can be exploited in a wide range of physical and mathematical applications.

One of the most useful consequences stems from the elementary observation that the com-position of two complex functions is also a complex function. We interpret this operationas a complex changes of variables, also known as a conformal mapping since it preserves an-gles. Conformal mappings can be effectively used for constructing solutions to the Laplaceequation on complicated planar domains that arise in a wide variety of physical problemsarising in fluid flow, aerodynamics, thermal equilibrium, electrostatics, elasticity, and soon.

In this chapter, we will develop the basic techniques and theorems of complex anal-ysis that impinge on the solution to boundary value problems associated with the planarLaplace and Poisson equations. We refer the reader to Appendix A for a quick review ofthe basics of complex numbers and complex arithmetic, and begin our exposition with thebasics of complex functions and the differential calculus. This allows us to develop thetheory and applications of conformal mappings in detail. The final section is devoted tocomplex integration and some of its basic applications.

6.1. Complex Functions.

Our main objects of study are complex-valued functions f(z), depending on a singlecomplex variable z = x + i y. In general, the function f: C will be defined on a

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subdomain z C of the complex plane.Any complex function can be uniquely written as a complex combination

f(z) = f(x + i y) = u(x, y) + i v(x, y), (6.1)

of two real functions, each depending on the two real variables x, y: its real part u(x, y) =

Re f(z) and its imaginary part v(x, y) = Im f(z). For example, according to the binomialformula, the monomial function f(z) = z3 is written as

z3 = (x + i y)3 = (x3 3 x y2) + i (3 x2y y3),and so

Re z3 = x3 3 x y2, Im z3 = 3 x2y y3.Many of the well-known functions appearing in real-variable calculus polynomials,

rational functions, exponentials, trigonometric functions, logarithms, and many others have natural complex extensions. For example, complex polynomials

p(z) = an zn

+ an1 zn1

+ + a1 z + a0 (6.2)are complex linear combinations (meaning that the coefficients ak are allowed to be complexnumbers) of the basic monomial functions zk = (x+ i y)k. Similarly, we have already madeuse of complex exponentials such as

ez = ex+ i y = ex cos y + i ex sin y,

when solving differential equations and in Fourier analysis. Further examples will appearshortly.

There are several ways to motivate the link between harmonic functions u(x, y), mean-ing solutions of the two-dimensional Laplace equation

u = 2u

x2+

2uy 2

= 0, (6.3)

and complex functions f(z). One natural starting point is to return to the dAlembertsolution (2.80) of the one-dimensional wave equation, which was based on the factorization

= 2t c2 2x = (t c x) (t + c x)of the linear wave operator (2.66). The two-dimensional Laplace operator = 2x +

2y

has essentially the same form, except for a minor change in sign. The Laplace operatoradmits a complex factorization,

= 2x +

2y = (x i y) (x + i y),

Including the Bessel functions, Airy functions and Legendre functions that we will encounterwhen we try to solve partial differential equations in more than two variables.

However, the change in sign has serious ramifications for the analytical properties of (real)solutions to the two equations. As noted in Section 4.3, there is a profound difference betweenthe elliptic Laplace equation and the hyperbolic wave equation.

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into a product of first order differential operators, with complex wave speeds c = i .Mimicking our previous solution formula (2.73) for the wave equation, we anticipate thatthe solutions to the Laplace equation (6.3) should be expressed in the form

u(x, y) = f(x + i y) + g(x i y), (6.4)i.e., a linear combination of functions of the complex variable z = x + i y and its complexconjugate z = x i y. The functions f(x + i y) and g(x i y) formally satisfy the firstorder complex partial differential equations

f

x= i f

y,

g

x= i

g

y, (6.5)

and hence (6.4) does indeed define a complex-valued solution to the Laplace equation.

In most applications, we are searching for a real solution to the Laplace equation, andso our complex dAlembert-type formula (6.4) is not entirely satisfactory. As we know, acomplex number z = x + i y is real if and only if it equals its own conjugate, z = z. Thus,

the solution (6.4) will be real if and only iff(x + i y) + g(x i y) = u(x, y) = u(x, y) = f(x + i y) + g(x i y).

Now, the complex conjugation operation interchanges x + i y and x i y, and so we expectthe first term f(x + i y) to be a function of x i y, while the second term g(x i y) willbe a function of x + i y. Therefore, to equate the two sides of this equation, we shouldrequire

g(x i y) = f(x + i y),and so

u(x, y) = f(x + i y) + f(x + i y) = 2 Re f(x + i y).

Dropping the inessential factor of 2, we conclude that a real solution to the two-dimensionalLaplace equation can be written as the real part of a complex function. A more directproof of the following key result will appear below.

Proposition 6.1. If f(z) is a complex function, then its real part

u(x, y) = Re f(x + i y) (6.6)

is a harmonic function.

The imaginary part of a complex function is also harmonic. This is because

Im f(z) = Re ( i f(z))is the real part of the complex function

i f(z) = i [ u(x, y) + i v(x, y)] = v(x, y) i u(x, y).

We are ignoring the fact that f and g are not quite uniquely determined since one can addand subtract a common constant. This does not affect the argument in any significant way.

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Re 1z Im1z

Figure 6.1. Real and Imaginary Parts of f(z) =1z .

Therefore, if f(z) is any complex function, we can write it as a complex combination

f(z) = f(x + i y) = u(x, y) + i v(x, y),

of two inter-related real harmonic functions: u(x, y) = Re f(z) and v(x, y) = Im f(z).

Before delving into the many remarkable properties of complex functions, let us lookat some of the most basic examples. In each case, the reader can directly check thatthe harmonic functions given as the real and imaginary parts of the complex function areindeed solutions to the Laplace equation.

Examples of Complex Functions

(a) Harmonic Polynomials: As noted above, any complex polynomial is a linear com-bination, as in (6.2), of the basic complex monomials

zn = (x + i y)n = un(x, y) + i vn(x, y). (6.7)

Its real and imaginary parts un, vn are the harmonic polynomials that we earlier con-structed by applying separation of variables to the polar coordinate form of the Laplaceequation (4.95). A general formula can be found in (4.111).

(b) Rational Functions: Ratios

f(z) =p(z)

q(z)

(6.8)

of complex polynomials provide a large variety of harmonic functions. The simplest caseis

1

z=

x

x2 + y2 i y

x2 + y2. (6.9)

Its real and imaginary parts are graphed in Figure 6.1. Note that these functions have aninteresting singularity at the origin x = y = 0, but are harmonic everywhere else.

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Figure 6.2. Real and Imaginary Parts of ez.

A slightly more complicated example is the function

f(z) =z 1z + 1

. (6.10)

To write out (6.10) in real form, we multiply and divide by the complex conjugate of thedenominator, leading to

f(z) =z 1z + 1

=(z 1)(z + 1)(z + 1)(z + 1)

=| z |2 + z z 1

| z + 1 |2 =x2 + y2 1

(x + 1)2 + y2+ i

2 y

(x + 1)2 + y2.

(6.11)Again, the real and imaginary parts are both harmonic functions away from the singularityx =

1, y = 0. Incidentally, the preceding manipulation can always be used to find the

real and imaginary parts of general rational functions.(c) Complex Exponentials : Eulers formula

ez = ex cos y + i ex sin y (6.12)

for the complex exponential yields two important harmonic functions: ex cos y and ex sin y,which are graphed in Figure 6.2. More generally, writing out ecz for a complex constantc = a + i b produces the complex exponential function

ecz = eaxby cos(b x + a y) + i eaxby sin(b x + a y). (6.13)

Its real and imaginary parts are harmonic functions for arbitrary a, b

R. Some of these

were found by applying the separation of variables method in Cartesian coordinates.(d) Complex Trigonometric Functions: The complex trigonometric functions are de-

fined in terms of the complex exponential by adapting our earlier formulae (3.60):

cos z =e i z + e i z

2= cos x cosh y i sin x sinh y,

sin z =e i z e i z

2 i= sin x cosh y + i cos x sinh y.

(6.14)

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Re (log z) = log | z | Im(log z) = ph zFigure 6.3. Real and Imaginary Parts of log z.

The resulting harmonic functions are products of trigonometric and hyperbolic functions.They can all be written as linear combinations of the harmonic functions (6.13) derivedfrom the complex exponential. Note that when z = x is real, so y = 0, these functionsreduce to the usual real trigonometric functions cos x and sin x.

(e) Complex Logarithm: In a similar fashion, the complex logarithm log z is a complexextension of the usual real natural (i.e., base e) logarithm. In terms of polar coordinatesz = r e i , the complex logarithm has the form

log z = log(r e i ) = log r + log e i = log r + i . (6.15)

Thus, the logarithm of a complex number has real part

Re (log z) = log r = log | z | = 12 log(x2 + y2),which is a well-defined harmonic function on all of R2 save for a logarithmic singularity atthe origin x = y = 0. It is, up to multiple, the logarithmic potential (5.107) correspondingto a delta function forcing concentrated at the origin a function that played a key rolein our construction of the Greens function for the Poisson equation.

The imaginary partIm(log z) = = ph z

of the complex logarithm is the phase (or argument) of z, also not defined at the originx = y = 0. Moreover, the phase is a multiply-valued harmonic function elsewhere, since itis only specified up to integer multiples of 2 . Thus, each nonzero complex number z = 0has an infinite number of possible values for its phase, and hence an infinite number ofpossible complex logarithms log z, differing from each other by an integer multiple of 2 i ,which reflects the fact that e2 i = 1. In particular, if z = x > 0 is real and positive, thenlog z = log x agrees with the real logarithm, provided we choose ph x = 0. Alternativechoices for the phase add in some integer multiple of 2 i , and so ordinary real, positive

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Figure 6.4. Real and Imaginary Parts of

z .

numbers x > 0 also have complex logarithms! On the other hand, if z = x < 0 is real andnegative, then log z = log | x | + (2 k + 1) i is complex no matter which value of ph z ischosen. (This explains why one avoids defining the logarithm of a negative number in firstyear calculus!)

Furthermore, as the point z circles once around the origin in a counter-clockwisedirection, Im log z = ph z = increases by 2 . Thus, the graph of ph z can be likened to aparking ramp with infinitely many levels, spiraling ever upwards as one circumambulatesthe origin; Figure 6.3 attempts to sketch it. At the origin, the complex logarithm exhibitsa type of singularity known as a logarithmic branch point, the branches referring to the

infinite number of possible values that can be assigned to log z at any nonzero point.(f) Roots and Fractional Powers : A similar branching phenomenon occurs with the

fractional powers and roots of complex numbers. The simplest case is the square rootfunction

z. Every nonzero complex number z = 0 has two different possible square

roots:

z and z. Writing z = r e i in polar coordinates, we find that

z =

r e i =

r e i /2 =

r

cos

2+ i sin

2

, (6.16)

i.e., we take the square root of the modulus and halve the phase:

z = | z | =

r , ph

z = 12 ph z =12 .

Since = ph z is only defined up to an integer multiple of 2 , the angle 12 is only definedup to an integer multiple of . The even and odd multiples account for the two possiblevalues of the square root.

In this case, if we start at some z = 0 and circle once around the origin, we increaseph z by 2 , but ph

z only increases by . Thus, at the end of our circuit, we arrive at the

other square root z. Circling the origin again increases ph z by a further 2 , and hencebrings us back to the original square root

z. Therefore, the graph of the multiply-valued

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square root function will look like a weirdly interconnected parking ramp with only twolevels, as sketched in Figure 6.4. The origin represents a branch point of order 2 for thesquare root function.

The preceding list of elementary examples is far from exhaustive. Lack of space will

preclude us from studying the remarkable properties of complex versions of the gammafunction, Airy functions, Bessel functions, and Legendre functions that appear later in thetext, as well as the Riemann zeta function (3.58), elliptic functions, modular functions,and many, many other important and fascinating functions arising in complex analysis andits manifold applications. The interested reader is referred to [131, 137].

6.2. Complex Differentiation.

The bedrock of complex function theory is the notion of the complex derivative. Com-plex differentiation is defined in the same manner as the usual calculus limit definition ofthe derivative of a real function. Yet, despite a superficial similarity, complex differentia-tion is a profoundly different theory, displaying an elegance and depth not shared by itsreal progenitor.

Definition 6.2. A complex function f(z) is differentiable at a point z C if andonly if the limiting difference quotient exists:

f(z) = limw z

f(w) f(z)w z . (6.17)

The key feature of this definition is that the limiting value f(z) of the differencequotient must be independent of how w converges to z. On the real line, there are onlytwo directions to approach a limiting point either from the left or from the right. These

lead to the concepts of left and right handed derivatives and their equality is required forthe existence of the usual derivative of a real function. In the complex plane, there arean infinite variety of directions to approach the point z, and the definition requires thatall of these directional derivatives must agree. This is the reason for the more severerestrictions on complex derivatives, and, in consequence, the source of their remarkableproperties.

Let us first see what happens when we approach z along the two simplest directions horizontal and vertical. If we set

w = z + h = (x + h) + i y, where h is real,

then w

z along a horizontal line as h

0, as sketched in Figure 6.5. If we write out

f(z) = u(x, y) + i v(x, y)

in terms of its real and imaginary parts, then we must have

f(z) = limh 0

f(z + h) f(z)h

= limh 0

f(x + h + i y) f(x + i y)h

= limh 0

u(x + h, y) u(x, y)

h+ i

v(x + h, y) v(x, y)h

=

u

x+ i

v

x=

f

x,

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z

z + h

z + i k

Figure 6.5. Complex Derivative Directions.

which follows from the usual definition of the (real) partial derivative. On the other hand,if we set

w = z + i k = x + i (y + k), where k is real,

then w z along a vertical line as k 0. Therefore, we must also have

f(z) = limk 0

f(z + i k) f(z)i k

= limk 0

i f(x + i (y + k)) f(x + i y)

k

= limh 0

v(x, y + k) v(x, y)

k i u(x, y + k) u(x, y)

k

=

v

y i u

y= i f

y.

When we equate the real and imaginary parts of these two distinct formulae for the complexderivative f(z), we discover that the real and imaginary components off(z) must satisfy acertain homogeneous linear system of partial differential equations, named after Augustin

Louis Cauchy and Bernhard Riemann, two of the founders of modern complex analysis.

Theorem 6.3. A function f(z) = u(x, y)+ i v(x, y), wherez = x+ i y, has a complexderivative f(z) if and only if its real and imaginary parts are continuously differentiableand satisfy the CauchyRiemann equations

u

x=

v

y,

u

y= v

x. (6.18)

In this case, the complex derivative of f(z) is equal to any of the following expressions:

f(z) =f

x=

u

x+ i

v

x= i f

y=

v

y i u

y. (6.19)

Cauchy also played an essential role in establishing the mathematics of elasticity and materialscience.

In addition to his contributions to complex analysis, partial differential equations and numbertheory, Riemann also was the inventor of the metric geometry of curved spaces, now known asRiemannian geometry, which turned out to be absolutely essential for Einsteins theory of generalrelativity some 70 years later!

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The proof of the converse that any function whose real and imaginary componentssatisfy the CauchyRiemann equations is differentiable will be omitted, but can befound in any basic text on complex analysis, e.g., [ 3, 113].

Remark: It is worth pointing out that equation (6.19) tells us that f satisfies f/x =

i f/y, which, reassuringly, agrees with the first equation in (6.5).Example 6.4. Consider the elementary function

z3 = (x3 3 x y2) + i (3x2y y3).Its real part u = x3 3 x y2 and imaginary part v = 3 x2y y3 satisfy the CauchyRiemannequations (6.18), since

u

x= 3 x2 3 y2 = v

y,

u

y= 6x y = v

x.

Theorem 6.3 implies that f(z) = z3 is complex differentiable. Not surprisingly, its deriva-tive turns out to be

f(z) =u

x+ i

v

x=

v

y i u

y= (3 x2 3 y2) + i (6 x y) = 3 z2.

Fortunately, the complex derivative obeys all of the usual rules that you learned inreal-variable calculus. For example,

d

dzzn = n zn1,

d

dzecz = c ecz,

d

dzlog z =

1

z, (6.20)

and so on. The power n can be non-integral or even, in view of the identity zn = en log z,complex, while c is any complex constant. The exponential formulae (6.14) for the complextrigonometric functions implies that they also satisfy the standard rules

d

dzcos z = sin z, d

dzsin z = cos z. (6.21)

The formulae for differentiating sums, products, ratios, inverses, and compositions of com-plex functions are all identical to their real counterparts, with similar proofs. Thus, thank-fully, you dont need to learn any new rules for performing complex differentiation!

Remark: There are many examples of seemingly reasonable functions which do nothave a complex derivative. The simplest is the complex conjugate function

f(z) = z = x i y.Its real and imaginary parts do not satisfy the CauchyRiemann equations, and hence zdoes not have a complex derivative. More generally, any function f(x, y) = h(z, z) thatexplicitly depends on the complex conjugate variable z is not complex-differentiable.

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Power Series and Analyticity

The most remarkable feature of complex analysis, which distinguishes it from realfunction theory, is that the existence of one complex derivative automatically implies theexistence of infinitely many! All complex functions f(z) are infinitely differentiable and,

in fact, analytic where defined. The reason for this surprising and profound fact will,however, not become evident until we learn the basics of complex integration in Section 6.6.In this section, we shall take analyticity as a given, and investigate some of its principalconsequences.

Definition 6.5. A complex function f(z) is called analytic at a point z0 if it has apower series expansion

f(z) = a0 + a1(z z0) + a2(z z0)2 + a3(z z0)3 + =n= 0

an (z z0)n , (6.22)

which converges for all z sufficiently close to z0.

Typically, the standard ratio or root tests for convergence of (real) series that youlearned in ordinary calculus, [7, 124], can be applied to determine where a given (complex)power series converges. We note that if f(z) and g(z) are analytic at a point z0, so is theirsum f(z) + g(z), product f(z) g(z) and, provided g(z0) = 0, ratio f(z)/g(z).

Example 6.6. All of the real power series found in elementary calculus carry overto the complex versions of the functions. For example,

ez = 1 + z + 12 z2 + 16 z

3 + =

n= 0zn

n!(6.23)

is the Taylor series for the exponential function based at z0 = 0. A simple applicationof the ratio test proves that the series converges for all z. On the other hand, the powerseries

1

z2 + 1= 1 z2 + z4 z6 + =

k= 0

(1)k z2k , (6.24)

converges inside the unit disk, where | z | < 1, and diverges outside, where | z | > 1. Again,convergence is established through the ratio test. The ratio test is inconclusive when| z | = 1, and we shall leave the more delicate question of precisely where on the unit diskthis complex series converges to a more advanced treatment, e.g., [ 3].

In general, there are three possible options for the domain of convergence of a complexpower series (6.22):

(a) The series converges for all z.

(b) The series converges inside a disk | z z0 | < of radius > 0 centered at z0 anddiverges for all | z z0 | > outside the disk. The series may converge at some(but not all) of the points on the boundary of the disk where | z z0 | = .

(c) The series only converges, trivially, at z = z0.

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The number is known as the radius of convergence of the series. In case (a), we say = , while in case (c), = 0, and the series does not represent an analytic function. Anexample with = 0 is the power series

n! zn. In the intermediate case (b), determining

precisely where on the boundary of the convergence disk the power series converges is quitedelicate, and will not be pursued here. The proof of this result can be found in Exercise

; see also [3, 63] for further details.Remarkably, the radius of convergence for the power series of a known analytic function

f(z) can be determined by inspection, without recourse to any fancy convergence tests!Namely, is equal to the distance from z0 to the nearest singularity of f(z), meaning apoint where the function fails to be analytic. This explains why the Taylor series of ez

converges everywhere, while that of (z2 + 1)1 only converges inside the unit disk. Indeedez is analytic for all z and has no singularities; therefore the radius of convergence of itspower series centered at any point z0 is equal to = . On the other hand, thefunction

f(z) =1

z2 + 1=

1

(z + i )(z

i )

has singularities at z = i , and so the series (6.24) has radius of convergence = 1,which is the distance from z0 = 0 to the singularities. Thus, the extension of the theoryof power series to the complex plane serves to explain the apparent mystery of why, asa real function, (1 + x2)1 is well-defined and analytic for all real x, but its power seriesonly converges on the interval (1, 1). It is the complex singularities that prevent itsconvergence when | x | > 1. If we expand (z2 + 1)1 in a power series at some other point,say z0 = 1 + 2 i , then we need to determine which singularity is closest. We compute| i z0 | = | 1 i | =

2, while | i z0 | = | 1 3 i | =

10, and so the radius of

convergence =

2 is the smaller. Thus we can determine the radius of convergencewithout any explicit formula for its (rather complicated) Taylor expansion at z0 = 1 + 2 i .

There are, in fact, only three possible types of singularities of a complex function f(z): Pole. A singular point z = z0 is called a pole of order n > 0 if and only if

f(z) =h(z)

(z z0)n, (6.25)

where h(z) is analytic at z = z0 and h(z0) = 0. The simplest example of such afunction is f(z) = a (z z0)n for a = 0 a complex constant.

Branch point. We have already encountered the two basic types: algebraic branchpoints, such as the function n

z at z0 = 0, and logarithmic branch points such as log z

at z0 = 0. The degree of the branch point is n in the first case and in the second. Essential singularity. By definition, a singularity is essential if it is not a pole or abranch point. Essential singularities are considerably more complicated than poles

and branch points. The quintessential example is the essential singularity at z0 = 0of the function e1/z.

Example 6.7. The complex function

f(z) =ez

z3 z2 5z 3 =ez

(z 3)(z + 1)2

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is analytic everywhere except for singularities at the points z = 3 and z = 1, where itsdenominator vanishes. Since

f(z) =h1(z)

z 3 , where h1(z) =ez

(z + 1)2

is analytic at z = 3 and h1(3) = 116 e3 = 0, we see that z = 3 is a simple (order 1) pole for

f(z). Similarly,

f(z) =h2(z)

(z + 1)2, where h2(z) =

ez

z 3is analytic at z = 1 with h2(1) = 14 e1 = 0, we see that the point z = 1 is a double(order 2) pole.

A complicated complex function can have a variety of singularities. For example, thefunction

f(z) =

3

z + 2 e1/z

z2 + 1 (6.26)

has simple poles at z = i , a branch point of degree 3 at z = 2, and an essentialsingularity at z = 0.

As in the real case, and unlike Fourier series, convergent power series can always berepeatedly term-wise differentiated. Therefore, given the convergent series (6.22), we havethe corresponding series

f(z) = a1 + 2 a2(z z0) + 3 a3(z z0)2 + 4 a4(z z0)3 +

=

n= 0 (n + 1) an+1(z z0)n,

f(z) = 2 a2 + 6 a3(z z0) + 12 a4(z z0)2 + 20 a5(z z0)3 +

=n= 0

(n + 1)(n + 2) an+2(z z0)n,

(6.27)

and so on, for its derivatives. The proof that the differentiated series have the sameradius of convergence can be found in [ 3, 113]. As a consequence, we deduce the followingimportant result.

Theorem 6.8. Any analytic function is infinitely differentiable.

In particular, when we substitute z = z0 into the successively differentiated series, wediscover that

a0 = f(z0), a1 = f(z0), a2 =

12

f(z0),

and, in general,

an =f(n)(z)

n!. (6.28)

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z0

Figure 6.6. Radius of Convergence.

Therefore, a convergent power series (6.22) is, inevitably, the usual Taylor series

f(z) =

n= 0

f(n)

(z0)n!

(z z0)n , (6.29)

for the function f(z) at the point z0.

Let us conclude this section by summarizing the fundamental theorem that character-izes complex functions. A complete, rigorous proof relies on complex integration theory,which is the topic of Section 6.6.

Theorem 6.9. Let C be an open set. The following properties are equivalent:(a) The function f(z) has a continuous complex derivative f(z) for all z .(b) The real and imaginary parts of f(z) have continuous partial derivatives and satisfy

the CauchyRiemann equations (6.18) in .(c) The function f(z) is analytic for all z , and so is infinitely differentiable and has a

convergent power series expansion at each point z0 . The radius of convergence is at least as large as the distance from z0 to the boundary ; see Figure 6.6.

From now on, we reserve the term complex function to signifiy one that satisfiesthe conditions of Theorem 6.9. Sometimes one of the equivalent adjectives analytic orholomorphic, is added for emphasis. From now on, all complex functions are assumed tobe analytic everywhere on their domain of definition, except, possibly, at certain isolatedsingularities.

6.3. Harmonic Functions.

We began this chapter by motivating the analysis of complex functions through ap-plications to the solution of the two-dimensional Laplace equation. Let us now formalizethe precise relationship between the two subjects.

Theorem 6.10. If f(z) = u(x, y) + i v(x, y) is any complex analytic function, thenits real and imaginary parts, u(x, y), v(x, y), are both harmonic functions.

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Proof: Differentiating the CauchyRiemann equations (6.18), and invoking theequality of mixed partial derivatives, we find that

2u

x2=

x

u

x

=

x

v

y

=

2v

xy=

y

v

x

=

y

u

y

=

2u

y 2.

Therefore, u is a solution to the Laplace equation uxx + uyy = 0. The proof for v issimilar. Q.E.D.

Thus, every complex function gives rise to two harmonic functions. It is, of course, ofinterest to know whether we can invert this procedure. Given a harmonic function u(x, y),does there exist a harmonic function v(x, y) such that f = u + i v is a complex analyticfunction? If so, the harmonic function v(x, y) is known as a harmonic conjugate to u. Theharmonic conjugate is found by solving the CauchyRiemann equations

v

x= u

y,

v

y=

u

x, (6.30)

which, for a prescribed function u(x, y), constitutes an inhomogeneous linear system of

partial differential equations for v(x, y). As such, it is usually not hard to solve, as thefollowing example illustrates.

Example 6.11. As the reader can verify, the harmonic polynomial

u(x, y) = x3 3 x2y 3 x y2 + y3satisfies the Laplace equation everywhere. To find a harmonic conjugate, we solve theCauchyRiemann equations (6.30). First of all,

v

x= u

y= 3 x2 + 6 x y 3 y2,

and hence, by direct integration with respect to x,

v(x, y) = x3 + 3 x2y 3x y2 + h(y),where h(y) the constant of integration is a function of y alone. To determine h wesubstitute our formula into the second CauchyRiemann equation:

3x2 6 x y + h(y) = vy

=u

x= 3 x2 6 x y 3 y2.

Therefore, h(y) = 3 y2, and so h(y) = y3 + c, where c is a real constant. We concludethat every harmonic conjugate to u(x, y) has the form

v(x, y) = x3 + 3 x2y 3 x y2 y3 + c.Note that the corresponding complex function

u(x, y) + i v(x, y) = (x3 3 x2y 3 x y2 + y3) + i (x3 + 3 x2y 3 x y2 y3 + c)= (1 i )z3 + c

turns out to be a complex cubic polynomial.

Theorem 6.9 allows us to differentiate u and v as often as desired.

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Remark: On a connected domain R2, all harmonic conjugates to a given functionu(x, y) only differ by a constant: v(x, y) = v(x, y) + c; see Exercise .

Although most harmonic functions have harmonic conjugates, unfortunately this isnot always the case. Interestingly, the existence or non-existence of a harmonic conjugate

can depend on the underlying topology of its domain of definition. If the domain is simplyconnected, and so contains no holes, then one can always find a harmonic conjugate. Onnon-simply connected domains, there may not exist a single-valued harmonic conjugate toserve as the imaginary part of a complex function f(z).

Example 6.12. The simplest example where the latter possibility occurs is thelogarithmic potential

u(x, y) = log r = 12 log(x2 + y2).

This function is harmonic on the non-simply connected domain = C \ {0}, but is notthe real part of any single-valued complex function. Indeed, according to (6.15), thelogarithmic potential is the real part of the multiply-valued complex logarithm log z, andso its harmonic conjugate is ph z = , which cannot be consistently and continuouslydefined on all of . On the other hand, on any simply connected subdomain , onecan select a continuous, single-valued branch of the angle = ph z, which is then a bonafide harmonic conjugate to log r restricted to this subdomain.

The harmonic function

u(x, y) =x

x2 + y2

is also defined on the same non-simply connected domain = C \ {0} with a singularityat x = y = 0. In this case, there is a single valued harmonic conjugate, namely

v(x, y) = y

x2 + y2 ,

which is defined on all of . Indeed, according to (6.9), these functions define the realand imaginary parts of the complex function u + i v = 1/z. Alternatively, one can directlycheck that they satisfy the CauchyRiemann equations (6.18).

Remark: On the punctured plane = C \ {0}, the logarithmic potential is, in asense, the only counterexample that prevents a harmonic conjugate from being constructed.It can be shown, [71], that if u(x, y) is a harmonic function defined on a punctured diskR =

0 < | z | < R

, where 0 < R , then there exists a constant c such that

u(x, y) = u(x, y) c logx2 + y2 is also harmonic and possess a single-valued harmonicconjugate v(x, y). As a result, the function f = u + i v is analytic on all of R, andso our original function u(x, y) is the real part of the multiply-valued analytic function

f(z) = f(z) + c log z. We shall use this fact in our later analysis of airfoils. We can, by a previous remark, add in any constant to the harmonic conjugate, but this does

not affect the subsequent argument.

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Theorem 6.13. Every harmonic function u(x, y) defined on a simply connecteddomain is the real part of a complex valued function f(z) = u(x, y) + i v(x, y) which isdefined for all z = x + i y .

Proof: We first rewrite the CauchyRiemann equations (6.30) in vectorial form as anequation for the gradient of v:

v = u, where u = uy, ux , (6.31)the skew gradient of u. As in (5.81), it is everywhere orthogonal to the gradient of u andof the same length:

u u = 0, u = u .Thus, we have established the important observation that the gradient of a harmonicfunction and that of its harmonic conjugate are mutually orthogonal vector fields havingthe same Euclidean lengths:

u

v

0,

u

v

. (6.32)

Now, given the harmonic function u, our goal is to construct a solution v to thegradient equation (6.31). A well-known result from vector calculus, [ 7], states the vectorfield defined by u has a potential function v if and only if the corresponding line integralis independent of path, which means that

0 =

C

v dx =C

u dx =C

u n ds, (6.33)

for every closed curve C . Indeed, if this holds, then a potential function can becomputed by integrating the vector field:

v(x, y) = xa

v dx = xa

u n ds. (6.34)Here a is any fixed point, and, in view of path independence, the line integral can be takenover any curve that connects a to x = (x, y).

If the domain is simply connected then every simple closed curve C bounds asudomain D with C = D. Applying the divergence form of Greens Theorem (5.83),we find

C

u n ds = D

u dxdy = D

u dxdy = 0,

because u is harmonic. Thus, we have proved the existence of a harmonic conjugatefunction provided the underlying domain is simply connected. Q.E.D.

We are continuing to use the convention of Section 5.3 that v = (vx, vy) is a row vector.

This assumes that the domain is connected; if not, we merely apply the argument on eachconnected component. Technically, we have only verified path-independence (6.33) when C is a simple closed curve,

but, by a standard argument, [7], this suffices to establish it for arbitrary closed curves.

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Figure 6.7. Level Curves of the Real and Imaginary Parts of z2 and z3.

Remark: As a consequence of (6.19) and the CauchyRiemann equations (6.30),

f

(z) =

u

x iu

y =

v

y + i

v

x . (6.35)Thus, the individual components of the gradients u and v appear as the real andimaginary parts of the complex derivative f(z).

The orthogonality (6.31) of the gradient of a function and of its harmonic conjugatehas the following important geometric consequence. Recall, [7, 124], that the gradient uof a function u(x, y) points in the normal direction to its level curves, that is, the sets{u(x, y) = c} where it assumes a fixed constant value. Since v is orthogonal to u, thismust mean that v is tangent to the level curves ofu. Vice versa, v is normal to its levelcurves, and so u is tangent to the level curves of its harmonic conjugate v. Since theirtangent directions

u and

v are orthogonal, the level curves of the real and imaginary

parts of a complex function form a mutually orthogonal system of plane curves butwith one key exception. If we are at a critical point, where u = 0, then v = u = 0,and the vectors do not define tangent directions. Therefore, the orthogonality of the levelcurves does not necessarily hold at critical points. It is worth pointing out that, in view of(6.35), the critical points of u are the same as those of v and also the same as the criticalpoints of the corresponding complex function f(z), i.e., those points where its complexderivative vanishes: f(z) = 0.

In Figure 6.7, we illustrate the preceding paragraph by plotting the level curves ofthe real and imaginary parts of the monomials z2 and z3. Note that, except at the origin,where the derivative vanishes, the level curves intersect everywhere at right angles.

Applications to Fluid Mechanics

Consider a planar steady state fluid flow, with velocity vector field

v(x) =

u(x, y)v(x, y)

at the point x = (x, y) .

Here R2 is the domain occupied by the fluid, while the vector v(x) represents theinstantaneous velocity of the fluid at the point x. Recall that the flow is incompressible if

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and only if it has vanishing divergence:

v = ux

+v

y= 0. (6.36)

Incompressibility means that the fluid volume does not change as it flows. Most liquids,

including water, are, for all practical purposes, incompressible. On the other hand, theflow is irrotational if and only if it has vanishing curl:

v = vx

uy

= 0. (6.37)

Irrotational flows have no vorticity, and hence no circulation. A flow that is both in-compressible and irrotational is known as an ideal fluid flow for short. In many physicalregimes, liquids (and, although less often, gases) behave as ideal fluids .

Observe that the two constraints (6.3637) are almost identical to the CauchyRiemannequations (6.18); the only difference is the change in sign in front of the derivatives of v.But this can be easily remedied by replacing v by its negative

v. As a result, we establish

a profound connection between ideal planar fluid flows and complex functions.

Theorem 6.14. The velocity vector field v = ( u(x, y), v(x, y) ) induces an ideal fluidflow if and only if

f(z) = u(x, y) i v(x, y) (6.38)is a complex analytic function of z = x + i y.

Thus, the components u(x, y) and v(x, y) of the velocity vector field for an idealfluid flow are necessarily harmonic conjugates. The corresponding complex function (6.38)is, not surprisingly, known as the complex velocity of the fluid flow. When using this result,do not forget the minus sign that appears in front of the imaginary part of f(z).

Under the flow induced by the velocity vector field v = ( u(x, y), v(x, y)), the fluidparticles follow the trajectories z(t) = x(t) + i y(t) obtained by integrating the system ofordinary differential equations

dx

dt= u(x, y),

dy

dt= v(x, y). (6.39)

In view of the representation (6.38), we can rewrite the system in complex form

dz

dt= f(z) . (6.40)

In fluid mechanics, the curves parametrized by z(t) are known as the streamlines. Each

fluid particles motion z(t) is uniquely prescribed by its position z(t0) = z0 = x0 + i y0 at aninitial time t0. In particular, if the complex velocity vanishes, f(z0) = 0, then the solutionz(t) z0 to (6.40) is constant, and hence z0 is a stagnation point of the flow. Our steadystate assumption, which is reflected in the fact that the ordinary differential equations(6.39) are autonomous, i.e., there is no explicit t dependence, means that, although thefluid is in motion, the stream lines and stagnation point do not change over time. This isa consequence of the standard existence and uniqueness theorems for solutions to ordinarydifferential equations [22, 61].

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f(z) = 1 f(z) = 4 + 3 i f(z) = z

Figure 6.8. Complex Fluid Flows.

Example 6.15. The simplest example is when the velocity is constant, correspond-ing to a uniform, steady flow. Consider first the case

f(z) = 1,which corresponds to the horizontal velocity vector field v = ( 1, 0 ). The actual fluid flowis found by integrating the system

z = 1, or

x = 1,

y = 0.

Thus, the solution z(t) = t+z0 represents a uniform horizontal fluid motion whose stream-lines are straight lines parallel to the real axis; see Figure 6.8.

Consider next a more general constant velocity

f(z) = c = a + i b.

The fluid particles will solve the ordinary differential equation

z = c = a i b, so that z(t) = c t + z0.The streamlines remain parallel straight lines, but now at an angle = ph c = ph c withthe horizontal. The fluid particles move along the streamlines at constant speed | c | = | c |.

The next simplest complex velocity function is

f(z) = z = x + i y. (6.41)

The corresponding fluid flow is found by integrating the system

z = z, or, in real form,

x = x,

y = y.The origin x = y = 0 is a stagnation point. The trajectories of the nonstationary solutions

z(t) = x0 et + i y0 e

t (6.42)

are the hyperbolas x y = c, along with the positive and negative coordinate semi-axes, asillustrated in Figure 6.8.

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Figure 6.9. Flow Inside a Corner.

On the other hand, if we choose

f(z) = i z = y i x,then the flow is the solution to

z = i z, or, in real form,

x = y,

y = x.

The solutions

z(t) = (x0 cosh t + y0 sinh t) + i (x0 sinh t + y0 cosh t),

move along the hyperbolas (and rays) x2 y2 = c2. Observe that this flow can be obtainedby rotating the preceding example by 45

.In general, a solid object in a fluid flow is characterized by the no-flux condition that

the fluid velocity v is everywhere tangent to the boundary, and hence no fluid flows intoor out of the object. As a result, the boundary will necessarily consist of streamlines andstagnation points of the idealized fluid flow. For example, the boundary of the upper rightquadrant Q = {x > 0, y > 0} C consists of the positive x and y axes (along with theorigin). Since these are streamlines of the flow with complex velocity (6.41), its restrictionto Q represents the flow past a 90 interior corner, which appears in Figure 6.9. Theindividual fluid particles move along hyperbolas as they flow past the corner.

Remark: We could also restrict this flow to the domain = C \ {x < 0, y < 0}consisting of three quadrants, corresponding to a 90

exterior corner. However, this flow isnot as physically relevant since it has an unrealistic asymptotic behavior at large distances.See Exercise for a more realistic flow around an exterior corner.

Now, suppose that the complex velocity f(z) admits a complex anti-derivative, i.e., acomplex analytic function

(z) = (x, y) + i (x, y) that satisfiesd

dz= f(z). (6.43)

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Using the formula (6.19) for the complex derivative,

d

dz=

x i

y= u i v, so

x= u,

y= v.

Thus,

= v, and hence the real part (x, y) of the complex function (z) defines a

velocity potential for the fluid flow. For this reason, the anti-derivative (z) is known as acomplex potential function for the given fluid velocity field.

Since the complex potential is analytic, its real part the potential function isharmonic, and therefore satisfies the Laplace equation = 0. Conversely, any harmonicfunction can be viewed as the potential function for some fluid flow. The real fluid velocityis its gradient v = . The harmonic conjugate (x, y) to the velocity potential alsoplays an important role, and, in fluid mechanics, is known as the stream function. It alsosatisfies the Laplace equation = 0, and the potential and stream function are relatedby the CauchyRiemann equations (6.18):

x = u =

y ,

y = v =

x . (6.44)

The level sets of the velocity potential, {(x, y) = c}, where c R is fixed, are knownas equipotential curves. The velocity vector v = points in the normal direction to theequipotentials. On the other hand, as we noted above, v = is tangent to the levelcurves {(x, y) = d} of its harmonic conjugate stream function. But v is the velocityfield, and so tangent to the streamlines followed by the fluid particles. Thus, these twosystems of curves must coincide, and we infer that the level curves of the stream functionare the streamlines of the flow, whence its name! Summarizing, for an ideal fluid flow,the equipotentials { = c} and streamlines { = d} form mutually orthogonal systemsof plane curves. The fluid velocity v =

is tangent to the stream lines and normal

to the equipotentials, whereas the gradient of the stream function is tangent to theequipotentials and normal to the streamlines.

The discussion in the preceding paragraph implicitly relied on the fact that the velocityis nonzero, v = = 0, which means we are not at a stagnation point, where the fluidis not moving. While streamlines and equipotentials might begin or end at a stagnationpoint, there is no guarantee, and, indeed, in general it is not the case that they meet atmutually orthogonal directions there.

Example 6.16. The simplest example of a complex potential function is

(z) = z = x + i y.

Thus, the velocity potential is (x, y) = x, while its harmonic conjugate stream functionis (x, y) = y. The complex derivative of the potential is the complex velocity,

f(z) =d

dz= 1,

which corresponds to the uniform horizontal fluid motion considered first in Example 6.15.Note that the horizontal stream lines coincide with the level sets {y = d} of the stream

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Figure 6.10. Equipotentials and Streamlines for (z) = z.

Figure 6.11. Equipotentials and Streamlines for (z) = 12

z2.

function, whereas the equipotentials {x = c} are the orthogonal system of vertical lines;see Figure 6.10.

Next, consider the complex potential function

(z) = 12

z2 = 12

(x2 y2) + i x y.The associated complex velocity

f(z) = (z) = z = x + i y

leads to the hyperbolic flow (6.42). The hyperbolic streamlines x y = d are the level curvesof the stream function (x, y) = x y. The equipotential lines 1

2(x2 y2) = c form a system

of orthogonal hyperbolas. Figure 6.11 shows (some of) the equipotentials in the first plot,the stream lines in the second, and combines them together in the third picture.

Example 6.17. Flow Around a Disk. Consider the complex potential function

(z) = z +1

z=

x +

x

x2 + y2

+ i

y y

x2 + y2

. (6.45)

The corresponding complex fluid velocity is

f(z) =d

dz= 1 1

z2= 1 x

2 y2(x2 + y2)2

+ i2x y

(x2 + y2)2. (6.46)

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Figure 6.12. Equipotentials and Streamlines for z +1

z.

Figure 6.13. Flow Past a Solid Disk.

The equipotential curves and streamlines are plotted in Figure 6.12. The points z = 1are stagnation points of the flow, while z = 0 is a singularity. In particular, fluid particlesthat move along the positive x axis approach the leading stagnation point z = 1 as t .Note that the streamlines

(x, y) = y yx2 + y2

= d

are asymptotically horizontal at large distances, and hence, far away from the origin, theflow is indistinguishable from uniform horizontal motion with complex velocity f(z) 1.

The level curve for the particular value d = 0 consists of the unit circle | z | = 1 andthe real axis y = 0. In particular, the unit circle | z | = 1 consists of two semicircular streamlines combined with the two stagnation points. The flow velocity vector field v = iseverywhere tangent to the unit circle, and hence satisfies the no flux condition along theboundary of the unit disk. Thus, we can interpret (6.46), when restricted to the domain =

| z | > 1, as the complex velocity of a uniformly moving fluid around the outsideof a solid circular disk of radius 1, as illustrated in Figure 6.13. In three dimensions, thiswould correspond to the steady flow of a fluid around a solid cylinder.

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D

g

Figure 6.14. Mapping to the Unit Disk.

Remark: In this section, we have focused on the fluid mechanical roles of a harmonicfunction and its conjugate. An analogous interpretation applies when (x, y) representsan electromagnetic potential function; the level curves of its harmonic conjugate (x, y)are the paths followed by charged particles under the electromotive force field v = .Similarly, if(x, y) represents the equilibrium temperature distribution in a planar domain,its level lines represent the isotherms curves of constant temperature, while the level

lines of its harmonic conjugate are the curves along which heat energy flows. Finally, if(x, y) represents the height of a deformed membrane, then its level curves are the contourlines of elevation. The level curves of its harmonic conjugate are the curves of steepestdescent along the membrane, i.e., the routes followed by, say, water flowing down themembrane.

6.4. Conformal Mapping.

As we now know, complex functions provide an almost inexhaustible supply of har-monic functions, i.e., solutions to the the two-dimensional Laplace equation. Thus, tosolve an associated boundary value problem, we merely find the complex function whose

real part matches the prescribed boundary conditions. Unfortunately, even for relativelysimple domains, this remains a daunting task.

The one case where we do have an explicit solution is that of a circular disk, where thePoisson integral formula (4.117) provides a complete solution to the Dirichlet boundaryvalue problem. (See also Exercise for the Neumann problem.) Thus, one evident solutionstrategy for the corresponding boundary value problem on a more complicated domain isto transform it into the solved case by an inspired change of variables.

Analytic Maps

The intimate connections between complex analysis and solutions to the Laplace equa-

tion inspires us to look at changes of variables defined by complex functions. To this end,we will re-interpret a complex analytic function

= g(z) or + i = p(x, y) + i q(x, y) (6.47)

as a mapping that takes a point z = x + i y belonging to a prescribed domain C to apoint = + i belonging to the image domain D = g() C. In many cases, the imagedomain D is the unit disk, as in Figure 6.14, but the method can also be applied to moregeneral domains. In order to unambigouously relate functions on to functions on D, we

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require that the analytic mapping (6.47) be one-to-one so that each point D comesfrom a unique point z . As a result, the inverse function z = g1() is a well-definedmap from D back to , which we assume is also analytic on all of D. The calculus formulafor the derivative of the inverse function

d

d g1() =

1

g(z) at = g(z), (6.48)

which remains valid for complex functions, implies that the derivative of g(z) must benonzero everywhere in order that g1() be differentiable. This condition,

g(z) = 0 at every point z , (6.49)will play a crucial role in the development of the method. Finally, in order to matchthe boundary conditions, we will assume that the mapping extends continuously to theboundary and maps it, one-to-one, to the boundary D of the image domain.

Before trying to apply this idea to solve boundary value problems for the Laplaceequation, let us look at some of the most basic examples of analytic mappings.

Example 6.18. The simplest nontrivial analytic maps are the translations

= z + = (x + a) + i (y + b), (6.50)

where = a + i b is a fixed complex number. The effect of (6.50) is to translate the entirecomplex plane in the direction given by the vector ( a, b ). In particular, the translationmaps the disk = {| z + | < 1} of radius 1 and center at the point to the unit diskD = {| | < 1}.

Example 6.19. There are two types of linear analytic maps. First are the scalingmaps

= z = x + i y, (6.51)

where = 0 is a fixed nonzero real number. This maps the disk | z | < 1/| | to the unitdisk | | < 1. Second are the rotations

= e i z = (x cos y sin ) + i (x sin + y cos ) (6.52)around the origin by a fixed (real) angle . This maps the unit disk to itself.

Example 6.20. Any non-constant affine transformation

= z + , = 0, (6.53)defines an invertible analytic map on all of C, whose inverse z = 1( ) is also affine.Writing = e i in polar coordinates, we see that the affine map (6.53) can be viewed asthe composition of a rotation (6.52), followed by a scaling (6.51), followed by a translation(6.50). As such, it takes the disk | z + | < 1 of radius 1/| | = 1/| | and center /to the unit disk | | < 1.

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Figure 6.15. The mapping = ez .

Example 6.21. A more interesting example is the complex function

= g(z) =1

z, or =

x

x2 + y2, = y

x2 + y2, (6.54)

which defines an inversion of the complex plane. The inversion is a one-to-one analyticmap everywhere except at the origin z = 0; indeed g(z) is its own inverse: g1() = 1/.Since g(z) = 1/z2 is never zero, the derivative condition (6.49) is satisfied everywhere.Note that | | = 1/| z |, while ph = ph z. Thus, if = | z | > denotes the exteriorof the circle of radius , then the image points = 1/z satisfy | | = 1/| z |, and hence theimage domain is the punctured disk D =

0 < | | < 1/. In particular, the inversion

maps the outside of the unit disk to its inside, but with the origin removed, and viceversa. The reader may enjoy seeing what the inversion does to other domains, e.g., theunit square S = { z = x + i y | 0 < x, y < 1 }.

Example 6.22. The complex exponential

= g(z) = ez , or = ex cos y, = ex sin y, (6.55)

satisfies the conditiong(z) = ez = 0

everywhere. Nevertheless, it is not one-to-one because ez+2 i = ez, and so all pointsdiffering by an integer multiple of 2 i are mapped to the same point. Thus, (6.49) isnecessary, but not sufficient for invertibility.

Under the exponential map, the horizontal line Im z = b is mapped to the curve = ex+ i b = ex(cos b + i sin b), which, as x varies from to , traces out the rayemanating from the origin that makes an angle ph = b with the real axis. Therefore, theexponential map will map a horizontal strip

Sa,b = a < Im z < b to a wedge-shaped domain a,b = a < ph < b,and is one-to-one provided | b a | < 2 . In particular, the horizontal strip

S/2,/2 = 1

2 < Im z 0,

while the horizontal strip S, = < Im z <

of width 2 is mapped onto the

domain = , = < ph < = C \ { Im z = 0, Re z 0}

obtained by cutting the complex plane along the negative real axis.

On the other hand, vertical lines Re z = a are mapped to circles | | = ea. Thus,a vertical strip a < Re z < b is mapped to an annulus ea < | | < eb, albeit many-to-one, since the strip is effectively wrapped around and around the annulus. The rectangleR =

a < x < b, < y < of height 2 is mapped in a one-to-one fashion on an

annulus that has been cut along the negative real axis. See Figure 6.15. Finally, we notethat no domain is mapped to the unit disk D = {| | < 1} (or, indeed, any other domainthat contains 0) because the exponential function is never zero: = ez = 0.

Example 6.23. The squaring map

= g(z) = z2, or = x2 y2, = 2 x y, (6.56)is analytic on all of C, but is not one-to-one. Its inverse is the square root functionz =

, which, as we noted in Section 6.1, is doubly-valued, except at the origin z =

0. Furthermore, the derivative g(z) = 2z vanishes at z = 0, violating the invertibilitycondition (6.49). However, once we restrict g(z) to a simply connected subdomain thatdoes not contain 0, the function g(z) = z2 does define a one-to-one mapping, whose inversez = g1() =

is a well-defined, analytic and single-valued branch of the square root

function.

The effect of the squaring map on a point z is to square its modulus, | | = | z |2

, whiledoubling its phase, ph = ph z2 = 2 ph z. Thus, for example, the upper right quadrant

Q =

x > 0, y > 0

=

0 < ph z < 12

is mapped onto the upper half plane

U = g(Q) =

= Im > 0

=

0 < ph <

.

The inverse function maps a point U back to its unique square root z = that liesin the quadrant Q. Similarly, a quarter disk

Q = 0 < | z | < , 0 < ph z 0of radius 2. On the other hand, the unit square S =

0 < x < 1, 0 < y < 1

is mapped

to a curvilinear triangular domain, as indicated in Figure 6.16; the edges of the square onthe real and imaginary axes map to the two halves of the straight base of the triangle,while the other two edges become its curved sides.

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Figure 6.16. The Effect of = z2 on Various Domains.

Example 6.24. A particularly important example is the analytic map

=z 1z + 1

=x2 + y2 1

(x + 1)2 + y2+ i

2 y

(x + 1)2 + y2, (6.57)

where we established the formulae for its real and imaginary parts in (6.11). The map isone-to-one with analytic inverse

z =1 +

1 =1 2 2

(1 )2 + 2 + i2

(1 )2 + 2 , (6.58)

provided z = 1 and = 1. This particular analytic map has the important propertyof mapping the right half plane R =

x = Re z > 0

to the unit disk D =

| |2 < 1.Indeed, by (6.58)

| |2 = 2 + 2 < 1 if and only if x = 1 2 2

(1 )2 + 2 > 0.

Note that the denominator does not vanish on the interior of the disk D.

The complex functions (6.53, 54, 57) are particular examples of linear fractional trans-formations

=

z +

z + , (6.59)

which form one of the most important classes of analytic maps. Here , , , are arbitrarycomplex constants, subject to the restriction

= 0,since otherwise (6.59) reduces to a trivial constant (and non-invertible) map. (Why?)

Example 6.25. The linear fractional transformation

=z

z

1, with | | < 1, (6.60)

maps the unit disk to itself, moving the origin z = 0 to the point = . To prove this, wenote that

| z |2 = (z )(z ) = | z |2 z z + | |2,| z 1 |2 = ( z 1)( z 1) = | |2 | z |2 z z + 1.

Subtracting these two formulae,

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f

Figure 6.17. A Conformal Map.

Thus, | z | < | z 1 |, which implies that

| | = | z || z 1 | < 1 provided | z | < 1, | | < 1,

and hence, as promised, lies within the unit disk.

The rotations (6.52) also map the unit disk to itself, while leaving the origin fixed. Itcan be proved, [3, 113], that the only invertible analytic mappings that take the unit diskto itself are obtained by composing such a linear fractional transformation with a rotation.

Proposition 6.26. If = g(z) is a one-to-one analytic map that takes the unit diskto itself, then

g(z) = e iz

z 1 for some | | < 1, 0 < 2 . (6.61)

Additional properties of linear fractional transformations are outlined in the exercises.

Conformality

A remarkable geometrical property enjoyed by all complex analytic functions is that,at non-critical points, they preserve angles, and therefore define conformal mappings. Con-formality makes sense for any inner product space, although in practice one usually dealswith Euclidean space equipped with the standard dot product.

Definition 6.27. A function g: Rn Rn is called conformal if it preserves angles.But what does it mean to preserve angles? In the Euclidean norm, the angle between

two vectors is defined by their dot product. However, most analytic maps are nonlinear,

and so will not map vectors to vectors since they will typically map straight lines to curves.However, if we interpret angle to mean the angle between two curves, as illustrated inFigure 6.17, then we can make sense of the conformality requirement. Thus, in order torealize complex functions as conformal maps, we first need to understand their effect oncurves.

Or, more precisely, their tangent vectors at the point of intersection; see below for details.

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ph

z

z

Figure 6.18. Complex Curve and Tangent.

In general, a curve C C in the complex plane is parametrized by a complex-valuedfunction

z(t) = x(t) + i y(t), a t b, (6.62)that depends on a real parameter t. Note that there is no essential difference between a

complex plane curve (6.62) and a real plane curve; we have merely switched from vectornotation x(t) = ( x(t), y(t) ) to complex notation z(t) = x(t)+ i y(t). All the usual vectorialcurve terminology (closed, simple, piecewise smooth, etc.) is used without any modificationhere. In particular, the tangent vector to the curve can be identified as the complex number

z(t) =

x(t)+ i

y(t), where we use dots to indicated derivatives with respect to the parametert. Smoothness of the curve is guaranteed by the requirement that

z(t) = 0.Example 6.28. (a) The curve

z(t) = e i t = cos t + i sin t, for 0 t 2 ,parametrizes the unit circle | z | = 1 in the complex plane. Its complex tangent

z(t) = i e i t = i z(t) is obtained by rotating z through 90.

(b) The complex curve

z(t) = cosh t + i sinh t =1 + i

2et +

1 i2

et, < t < ,parametrizes the right hand branch of the hyperbola Re z2 = x2 y2 = 1. Thecomplex tangent vector is

z(t) = sinh t + i cosh t = i z(t).

When we interpret the curve as the motion of a particle in the complex plane, so thatz(t) is the position of the particle at time t, the tangent

z(t) represents its instantaneous

velocity. The modulus of the tangent, | z | =

x2 +

y2, indicates the particles speed,while its phase ph

z measures the direction of motion, as measured by the angle that the

curve makes with the horizontal; see Figure 6.18.The angle between between two curves is defined as the angle between their tangents

at the point of intersection. If the curve C1 has angle 1 = ph

z1(t1) while the curve C2 hasangle 2 = ph

z2(t2) at the common point z = z1(t1) = z2(t2), then the angle betweenC1 and C2 at z is their difference

= 2 1 = ph

z2 ph

z1 = ph

z2

z1

. (6.63)

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Now, consider the effect of an analytic map = g(z). A curve C parametrized by z(t)will be mapped to a new curve = g(C) parametrized by the composition (t) = g(z(t)).The tangent to the image curve is related to that of the original curve by the chain rule:

d

dt

=dg

dz

dz

dt

, or

(t) = g(z(t))

z(t). (6.64)

Therefore, the effect of the analytic map on the tangent vector

z is to multiply it by thecomplex number g(z). If the analytic map satisfies our key assumption g(z) = 0, then

= 0, and so the image curve is guaranteed to be smooth. Otherwise, it may have acorner, cusp, or other type of singularity.

According to equation (6.64),

|

| = | g(z) z | = | g(z) | | z |. (6.65)Thus, the speed of motion along the new curve (t) is multiplied by a factor = | g(z) | > 0.The magnification factor depends only upon the point z and not how the curve passes

through it. All curves passing through the point z are speeded up (or slowed down if < 1)by the same factor! Similarly, the angle that the new curve makes with the horizontal isgiven by

ph

= ph

g(z)

z

= ph g(z) + ph

z, (6.66)

since the phase of the product of two complex numbers is the sum of their individualphases. Therefore, the tangent angle of the curve is increased by an amount = ph g(z),which means that the tangent is been rotated through angle . Again, the increase intangent angle only depends on the point z, and all curves passing through z are rotated bythe same amount . As an immediate consequence, the angle between any two curves ispreserved. More precisely, ifC1 is at angle 1 and C2 at angle 2 at a point of intersection,

then their images 1 = g(C1) and 2 = g(C2) are at angles 1 = 1 + and 2 = 2 + .The angle between the two image curves is the difference

2 1 = (2 + ) (1 + ) = 2 1,which is the same as the angle between the original curves. This establishes the confor-mality or angle-preservation property of analytic maps.

Theorem 6.29. If = g(z) is an analytic function and g(z) = 0, then g defines aconformal map.

Remark: The converse is also valid: Every planar conformal map comes from a com-

plex analytic function with nonvanishing derivative. A proof is outlined in Exercise .

The conformality of analytic functions is all the more surprising when one revisitselementary examples. In Example 6.23, we discovered that the function w = z2 mapsa quarter plane to a half plane, and therefore doubles the angle between the coordinateaxes at the origin! Thus g(z) = z2 is most definitely not conformal at z = 0. Theexplanation is, of course, that z = 0 is a critical point, g(0) = 0, and Theorem 6.29 onlyguarantees conformality when the derivative is nonzero. Amazingly, the map preserves

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Figure 6.19. Conformality of z2.

Figure 6.20. The Joukowski Map.

angles everywhere else! Somehow, the angle at the origin is doubled, while angles at allnearby points are preserved. Figure 6.19 illustrates this remarkable and counter-intuitivefeat. The left hand figure shows the coordinate grid, while on the right are the images ofthe horizontal and vertical lines under the map z2. Note that, except at the origin, theimage curves continue to meet at 90 angles, in accordance with conformality.

Example 6.30. A particularly interesting conformal transformation is given by theJoukowski map

=1

2 z +1

z . (6.67)It is used in the study of flows around airplane wings, and named after the pioneeringRussian aero- and hydro-dynamics researcher Nikolai Zhukovskii (Joukowski). Since

d

dz=

1

2

1 1

z2

= 0 if and only if z = 1,

the Joukowski map is conformal except at the critical points z = 1, as well as at thesingularity z = 0 where it is not defined.

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Center: .1Radius: .5

Center: .2 + iRadius: 1

Center: 1 + iRadius: 1

Center: 2 + 3 iRadius: 3

2 4.2426

Center: .2 + iRadius: 1.2806

Center: .1 + .3 iRadius: .9487

Center: .1 + .1 iRadius: 1.1045

Center: .2 + .1 iRadius: 1.2042

Figure 6.21. Airfoils Obtained from Circles via the Joukowski Map.

If z = e i lies on the unit circle, then

= 12

e i + e i

= cos ,

lies on the real axis, with 1 1. Thus, the Joukowski map squashes the unit circledown to the real line segment [ 1, 1]. The images of points outside the unit circle fill therest of the plane, as do the images of the (nonzero) points inside the unit circle. Indeed,

if we solve (6.67) forz = 2 1 , (6.68)

we see that every except 1 comes from two different points z; for not on the criticalline segment [1, 1], one point lies inside and and one lies outside the unit circle, whereasif1 < < 1, the points are situated directly above and below it on the circle. Therefore,(6.67) defines a one-to-one conformal map from the exterior of the unit circle

| z | > 1 onto the exterior of the unit line segment C \ [1, 1 ].

Under the Joukowski map, the concentric circles | z | = r = 1 are mapped to ellipseswith foci at 1 in the plane; see Figure 6.20. The effect on circles not centered at theorigin is quite interesting. The image curves take on a wide variety of shapes; several

examples are plotted in Figure 6.21. If the circle passes through the singular point z = 1,then its image is no longer smooth, but has a cusp at = 1; this happens in the last 6of the figures. Some of the image curves have the shape of the cross-section through anairplane wing or airfoil. Later we will see how to construct the physical fluid flow aroundsuch an airfoil, a result that was a critical step in early aircraft design.

Composition and the Riemann Mapping Theorem

One of the strengths of the method of conformal mapping is that one can assemble a

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large repertoire of complicated examples by simply composing elementary mappings. Themethod rests on the simple fact that the composition of two complex analytic functions isalso complex analytic. This is the complex counterpart of the result, learned in first yearcalculus, that the composition of two differentiable functions is itself differentiable.

Proposition 6.31. If w = f(z) is an analytic function of the complex variablez = x + i y, and = g(w) is an analytic function of the complex variable w = u + i v, thenthe composition = h(z) g f(z) = g(f(z)) is an analytic function of z.

Proof: The proof that the composition of two differentiable functions is differentiableis identical to the real variable version, [ 7, 124], and need not be reproduced here. Thederivative of the composition is explicitly given by the usual chain rule:

d

dzg f(z) = g(f(z)) f(z), or, in Leibnizian notation,

d

dz=

d

dw

dw

dz. (6.69)

Further details are left to the reader. Q.E.D.

If both f and g are one-to-one, so is the composition h = g f. Moreover, the compo-

sition of two conformal maps is also conformal, a fact that is immediate from the definition,or by using the chain rule (6.69) to show that

h(z) = g(f(z)) f(z) = 0 provided g(f(z)) = 0 and f(z) = 0.Thus, if f and g satisfy the conformality condition (6.49), so does h = g f. Q.E.D.

Example 6.32. As we learned in Example 6.22, the exponential function

w = ez

maps the horizontal strip S = {12 < Im z < 12 } conformally onto the right half planeR = {Re w > 0}. On the other hand, Example 6.24 tells us that the linear fractionaltransformation = w 1

w + 1

maps the right half plane R conformally to the unit disk D = {| | < 1}. Therefore, thecomposition

=ez 1ez + 1

(6.70)

is a one-to-one conformal map from the horizontal strip S to the unit disk D, which weillustrate in Figure 6.22.

Recall that our motivating goal is to use analytic functions/conformal maps to solveboundary value problems for the Laplace equation on a complicated domain by trans-

forming them to boundary value problems on the unit disk. Of course, the key questionthe student should be asking at this point is: Is there, in fact, a conformal map = g(z)from a given domain to the unit disk D = g()? The theoretical answer is the celebratedRiemann Mapping Theorem.

Of course, to properly define the composition, we need to ensure that the range of the functionw = f(z) is contained in the domain of the function = g(w).

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w = ez =

w 1w + 1

Figure 6.22. Composition of Conformal Maps.

Theorem 6.33. If C is any simply connected open subset, not equal to theentire complex plane, then there exists a one-to-one complex analytic map = g(z),satisfying the conformality condition g(z) = 0 for all z , that maps to the unit diskD = {| | < 1}.

Thus, any simply connected domain with one exception, the entire complex plane can be conformally mapped the unit disk. Note that need not be bounded for

this to hold. Indeed, the conformal map (6.57) takes the unbounded right half planeR = {Re z > 0} to the unit disk. The proof of this important theorem relies on somemore advanced results in complex analysis, and can be found, for instance, in [ 3].

The Riemann Mapping Theorem guarantees the existence of a conformal map fromany simply connected domain to the unit disk, but its proof is not constructive, and so is oflittle help for constructing the desired mapping. And, in general, this is not an easy task.In practice, one assembles a repertoire of useful conformal maps that apply to particulardomains of interest. An extensive catalog can be found in [71]. More complicated mapscan then be built up by composition of the basic examples. Ultimately, though, thedetermination of a suitable conformal map is more an art than a systematic science.

Example 6.34. Suppose we are asked to conformally map the upper half planeU =

Im z > 0

to the unit disk D =

| | < 1 . We already know that the linearfractional transformation

= g(w) =w 1w + 1

maps the right half plane R =

Re z > 0

to D = g(R). On the other hand, multiplication

by i = e i/2, with z = h(w) = i w, rotates the complex plane by 90 and so maps theright half plane R to the upper half plane U = h(R). Its inverse h1(z) = i z willtherefore map U to R = h1(U). Therefore, to map the upper half plane to the unit disk,we compose these two maps, leading to the conformal map

= g h1(z) = i z 1 i z + 1 =

i z + 1

i z 1 (6.71)

from U to D.

In a similar vein, we already know that the squaring map w = z2 maps the upperright quadrant Q =

0 < ph z < 1

2

to the upper half plane U. Composing this withour previously constructed map which requires using w instead of z in the previous

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formula (6.71) leads to the conformal map

=i z2 + 1

i z2 1 (6.72)that maps the quadrant Q to the unit disk D.

Example 6.35. The goal of this example is to construct an conformal map thattakes a half disk

D+ = | z | < 1, y = Im z > 0 (6.73)

to the full unit disk D = | | < 1. The answer is not = z2 because the im-

age of D+ omits the positive real axis, resulting in a disk with a slit cut out of it: | | < 1, 0 < ph < 2 . To obtain the entire disk as the image of the conformal map,we must think a little harder. The first observation is that the map z = (w 1)/(w + 1)that we analyzed in Example 6.24 takes the right half plane R =

Re w > 0

to the unit

disk. Moreover, it maps the upper right quadrant Q =

0 < ph w < 1

2

to the half disk

(6.73). Its inverse,

w = z + 1z 1 (6.74)

will therefore map the half disk, z D+, to the upper right quadrant w Q.On the other hand, we just constructed a conformal map (6.72) that takes the upper

right quadrant Q to the unit disk D. Therefore, if compose the two maps (replacing z byw in (6.72) and then using (6.74)), we obtain the desired conformal map:

=i w2 + 1

i w2 1 =i

z + 1

z 12

+ 1

i z + 1z 1 2

1

=( i + 1)(z2 + 1) + 2( i 1)z( i 1)(z2 + 1) + 2( i + 1)z .

The formula can be further simplified by multiplying numerator and denominator by i +1,and so

= i z2 + 2 i z + 1

z2 2 i z + 1 .The leading factor i is unimportant and can be omitted, since it merely rotates the diskby 90, and so

=z2 + 2 i z + 1

z2 2 i z + 1 (6.75)is an equally valid solution to our problem.

Finally, as noted in the preceding example, the conformal map guaranteed by theRiemann Mapping Theorem is not unique. Since the linear fractional transformations(6.60) map the unit disk to itself, we can compose them with any conformal Riemannmapping to produce additional conformal maps from a simply connected domain to theunit disk. For example, composing (6.60) with (6.70) produces a family of mappings

=1 + ez (1 ez) (1 + ez) 1 + ez , (6.76)

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Figure 6.23. An Annulus.

which, for any | | < 1, maps the strip S = 12 < Im z < 12 onto the unit disk.Proposition 6.26 implies that this is the only ambiguity, and so, for instance, (6.76) formsa complete list of one-to-one conformal maps from S to D.

Annular Domains

The Riemann Mapping Theorem does not apply to non-simply connected domains.For purely topological reasons, a hole cannot be made to disappear under a one-to-onecontinuous mapping much less a conformal map and so a non-simply connecteddomain cannot be mapped in a one-to-one manner onto the unit disk. So we must lookelsewhere for a simple model domain.

The simplest non-simply connected domain is an annulus consisting of the pointsbetween two concentric circles

Ar,R =

r < | | < R, (6.77)which, for simplicity, is centered around the origin; see Figure 6.23. The case r = 0corresponds to a punctured disk, while setting R = gives the exterior of a disk of radiusr. It can be proved, [71], that any other domain with a single hole can be mapped to anannulus. The annular radii r, R are not uniquely specified; indeed the linear map = zmaps the annulus (6.77) to a rescaled annulus Ar,R whose inner and outer radii have

both been scaled by the factor = | |. But the ratio r/R of the inner to outer radius ofthe annulus is uniquely specified; annuli with different ratios cannot be mapped to eachother by a conformal map.

Example 6.36. Let c > 0. Consider the domain

= | z | < 1 and | z c | > c contained between two nonconcentric circles. To keep the computations simple, we takethe outer circle to have radius 1 (which can always be arranged by scaling, anyway) whilethe inner circle has center at the point z = c on the real axis and radius c, which means

If r = 0 or R = , but not both, then r/R = 0 by convention. The punctured plane, wherer = 0 and R = remains a separate case.

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Figure 6.24. Conformal Map for a Non-Concentric Annulus.

that it passes through the origin. We must restrict c < 12 in order that the inner circle not

overlap with the outer circle. Our goal is to find a conformal map = g(z) that takes thisnon-concentric annular domain to a concentric annulus of the form

Ar,1 =

r < | | < 1.Now, according, to Example 6.25, a linear fractional transformation of the form

= g(z) =z

z 1 with | | < 1 (6.78)

maps the unit disk to itself. Moreover, as remarked earlier, and demonstrated in Exercise, linear fractional transformations always map circles to circles. Therefore, we seek a

particular value of that maps the inner circle | z c | = c to a circle of the form | | = rcentered at the origin. We choose to be real and try to map the points 0 and 2 c on theinner circle to the points r and r on the circle | | = r. This requires

g(0) = = r, g(2 c) =2c

2 c 1 = r. (6.79)

Substituting the first into the second leads to the quadratic equation

c 2 + c = 0.There are two real solutions:

=1 1 4 c2

2 cand =

1 + 1 4 c22 c

. (6.80)

Since 0 < c < 12

, the second solution gives > 1, and hence is inadmissible. Therefore,the first solution yields the required conformal map

=z 1 + 1 4 c2

(1 1 4 c2 ) z 2 c .

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Note in particular that the radius r = of the inner circle in Ar,1 is not the same as

the radius c of the inner circle in . For example, taking c = 25 , equation (6.80) implies

= 12 , and hence the linear fractional transformation =2 z 1z 2 maps the annular domain

= | z | < 1, z 25 >

25 to the concentric annulus A = A1/2,1 =

12 0

to the unit disk D = | | < 1. Propo-

sition 6.37 implies that if U(, ) is a harmonic function in the unit disk, then

u(x, y) = U

x2 + y2 1

(x + 1)2 + y2,

2 y

(x + 1)2 + y2

(6.88)

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is a harmonic function on the right half plane. (This can, of course, be checked directlyby a rather unpleasant chain rule computation.)

To solve the Dirichlet boundary value problem

u = 0, x > 0, u(0, y) = h(y), (6.89)

on the right half plane, we adopt the change of variables (6.87) and use the Poisson integralformula to construct the solution to the transformed Dirichlet problem

U = 0, 2 + 2 < 1, U(cos , sin ) = H(), (6.90)

on the unit disk. The relevant boundary conditions are found as follows. Using the explicitform

x + i y = z =1 +

1 =

(1 + )(1 )| 1 |2

=1 + | |2

| 1 |2=

1 2 2 + 2 i ( 1)2 + 2

for the inverse map, we see that the boundary point = + i = e i on the unit circleD will correspond to the boundary point

i y =2

( 1)2 + 2 =2 i sin

(cos 1)2 + sin2 = i cot

2(6.91)

on the imaginary axis R =

Re z = 0

. Thus, the boundary data h(y) on R correspondsto the boundary data

H() = h cot 12 on the unit circle. The Poisson integral formula (4.117) can then be applied to solve (6.90),from which we are able to reconstruct the solution (6.88) to the boundary value problem(6.88) on the half plane.

Lets look at an explicit example. If the boundary data on the imaginary axis isprovided by the step function

u(0, y) = h(y)

1, y > 0,

0, y < 0,

then the corresponding boundary data on the unit disk is a (periodic) step function

H() =

1, 0 < < ,

0, < < 2 ,

that has values +1 on the upper semicircle, 0 on the lower semicircle, and jump discon-tinuities at = 1. According to the Poisson formula (4.117), the solution to the latter

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Figure 6.25. The Solution for the Disk and the Half Plane.

boundary value problem is given by

U(, ) =1

2

0

1 21 + 2 2 cos( ) d where

= cos ,

= sin ,

=

1

tan1

1 +

1 cot

2

+ tan1

1 +

1 tan

2

, 0 < < ,

1 +1

tan1

1 +

1 cot

2

+ tan1

1 +

1 tan

2

, < < 0,

0, =

,.

Finally, we use (6.88) to construct the solution on the right half plane, although we shallspare the reader the messy details of the final formula. The result is depicted in Figure 6.25.

Remark: The solution to the preceding Dirichlet boundary value problem is not, infact, unique, owing to the unboundedness of the domain. The solution that we pick outby using the conformal map to the unit disk is the one that remains bounded at . Theunbounded solutions would correspond to solutions on the unit disk that have a singularityin their boundary data at the point 1; see Exercise .

Example 6.39. A non-coaxial cable. The goal of this example is to determinethe electrostatic potential inside a non-coaxial cylindrical cable with prescribed constant

potential values on the two bounding cylinders, as illustrated in Figure 6.26. Assumefor definiteness that the larger cylinder has radius 1, and is centered at the origin, whilethe smaller cylinder has radius 25 , and is centered at z =

25 . The resulting electrostatic

potential will be independent of the longitudinal coordinate, and so can be viewed as aplanar potential in the annular domain contained between two circles representing the

We use the usual branch 12 < tan1 x < 12 of the inverse tangent.

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Figure 6.26. A NonCoaxial Cable.

cross-sections of our cyli

con formal map

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